× Didn't find what you were looking for? Ask a question
Top Posters
Since Sunday
w
5
a
3
j
2
a
2
t
2
u
2
r
2
j
2
j
2
l
2
d
2
y
2
New Topic  
riverside07 riverside07
wrote...
Posts: 63
Rep: 0 0
11 years ago
5.00 g of HCl are allowed to react with 25.0 g of zinc metal in the presence of water. During the reaction, zinc(II) chloride and hydrogen gas result. After the reaction is finished, [Zn2+(aq)] = 0.221 M, and the volume of the solution is 56.0 mL. What is the apparent percent yield of the reaction?
Read 395 times
1 Reply

Related Topics

Replies
wrote...
11 years ago
moles Zn = 25.0 g/ 65.39 g/mol=0.382
Zn + 2 HCl = ZnCl2 + H2
moles HCl required = 0.382 x 2 =0.764
moles HCl = 5.00 g /36.461 g/mol=0.137
HCl is the limiting reactant
moles Zn2+ formed = 0.137/2=0.0686
[Zn2+]= 0.0686/ 0.0560 L=1.22 M ( theoretical concentration)
% = 0.221 x 100/ 1.22 = 18.1
New Topic      
Explore
Post your homework questions and get free online help from our incredible volunteers
  1240 People Browsing
 217 Signed Up Today
Related Images
  
 260
  
 192
  
 328
Your Opinion
Which industry do you think artificial intelligence (AI) will impact the most?
Votes: 484

Previous poll results: What's your favorite math subject?